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Voltage drop through a transformer

  1. Nov 11, 2013 #1
    Hi, I understand that for a real transformer there will obviously be a current drop through the transformer due to winding resistance, core losses hysteresis etc. For the sake of power system analysis, however, do you still assume that the secondary and primary voltages are linked by terms of their turns ratio, or will the secondary voltage be the primary x turns ratio minus a small amount due to losses? Sorry if this is confusing
  2. jcsd
  3. Nov 12, 2013 #2

    The Electrician

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    More like this:

    The secondary voltage be the primary x turns ratio minus some due to leakage inductance minus a small amount due to losses.

    One might also account for lower than expected output as an effect of the coupling coefficient (qv).
  4. Nov 12, 2013 #3
    The magnetic flux flowing through the magnetic core produces EMF in both primary and secondary winding proportionally with number of turns.
    Part of magnetic flux of each winding is spreading in the surrounding medium-leakage flux- determining the Xp and Xs.
    If the current is low-as in no-load state when the secondary current is zero and the primary current is only 2-5% of rated you may neglect the voltage drop in primary resistance and reactance and consider E1/E2=Volt1/Volt2=n1/n2.
    Where E1,E2 =EMF in primary(1) and respective secondary(2) winding.
    Volt1,2=supply voltage (1), secondary voltage at the secondary terminals(2)
    n1,n2=primary[secondary] winding number of turns
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